RE: st: how to find the integral for a portion of a normal distribution.

["Buzz Burhans" <buzzb3@earthlink.net>]

Thanks for this response Steve.  In that paper they are interested in the
distribution of a different yield quantity than I am; they are looking at
the distribution of milk yield per milking.  Furthermore, there were a
couple of differences in the data they had available; for yield per milking
a negative yield can not occur, and they also truncated the lowest portion
of the yield data (4-17% of the observations).

My yield interest was a bit different. My interest was ultimately in
examining the economic risk to dairy farmers of implementing a feed additive
with a mean incremental response of 2.05 lb / day.  The mean increment of
yield response was determined in a meta-analysis commissioned from an
independent scientist by the manufacturer.  The assumption is then that the
increment of yield response is normally distributed across the approximately
50 trials in the analysis. (By the way, the meta-analysis was done using
Stata, was done by an excellent scientist, and is well done.)  The response
to such an intervention can also be negative (less than 0), and indeed was
in several of the trials included in the meta-analysis.

Too often, in the dairy industry (and everywhere else) decisions are
proposed and made based on a mean response, outcome, or performance; in
reality, the response to such interventions is not fixed, but occurs over a
distribution of possible responses.  This company is marketing the product
based strictly on the cost/return of the mean response.  I was initially
interested in the probability of seeing a breakeven response, and I was also
exploring the expected response conditional on being able to determine that
a herd would or would not see a response above the cutpoint of 1 lb.  

Thanks again for your response

Buzz



Buzz Burhans, Ph.D. 

Dairy-Tech Group
So. Albany, VT / Twin Falls ID

Phone: 802-755-6842
Cell: 208-320-0829
Fax VT: 802-755-6842
Fax ID: 208-735-1289

Email: buzzb3@earthlink.net

-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Steve Samuels
Sent: Tuesday, May 04, 2010 2:04 PM
To: statalist@hsphsun2.harvard.edu
Subject: Re: st: how to find the integral for a portion of a normal
distribution.

Buzz Burhans
My question here, technicalities aside, is why you think that the
normal distribution is a good fit to milk yields. A earlier post in
the thread already warned that the normal distribution with your
parameters has substantial probability below zero. This publication,
for what it's worth-I haven't read it, found a modified Weibull
distribution to be a good fit to milk
yields.((http://www.ncbi.nlm.nih.gov/pubmed/8046070) ) If you have
prior data, then -kdensity- will overlay the best fitting Normal on
top of a nonparametric estimated, and there are other diagnostic plots
in Stata (-help diagnostic plots-) You can fit parameters to the
Weibull and related distributions with -streg-. The Weibull is also
easy to simulate because there is a closed-form version of the
cumulative probability distribution.
Steve


On Tue, May 4, 2010 at 2:44 PM, Brian P. Poi <bpoi@stata.com> wrote:
> On Tue, 4 May 2010,
>
>>
>> Suppose I have an intervention applied to cows with a demonstrated mean
>> milk
>> yield response of +2.05 liters, sd 1.74.
>>
>> Suppose I am interested a 1 liter cut point.  I know how to find the
>> proportion of responses at or below the 1 liter response; and the
>> proportion
>> of responses at or above the one liter response.  This is what you are
>> doing
>> here, or it can be done with a z score.
>>
>> My question is, if the response is normally distributed, given n cows,
how
>> many total liters were included or accumulated in the only responses
below
>> 1
>> liter, and how many total liters were accumulated in only the responses
at
>> or above 1 liter.  (Negative responses are possible). My interest is in
>> the
>> total liters, not the fraction of the total population either above or
>> below
>> the cutpoint.
>>
>
> Assuming I understand the question correctly, I think this does what you
> want.
>
> For a random variable distributed N(mu, sigma),
>
>   E[X | X < 1] = mu + sigma*lambda(alpha)
>
> where
>
>   alpha = (1 - mu) / sigma
>   lambda(alpha) = -phi(alpha) / Phi(alpha)
>
> Given a sample of size n, the number of observations less than 1 is
>
>   numltone = n*Phi(alpha)
>
> so the total amount of milk given that X < 1 is
>
>   milk = numltone*E[X | X < 1]
>
> In Stata,
>
> . clear all
> . set mem 200m
> (204800k)
> . set seed 1
> . drawnorm x, means(2.05) sds(1.74) n(10000000) clear
> (obs 10000000)
> . summ x if x < 1, mean
> . di r(sum)
> -188292.37
>
> . scalar alpha = (1 - 2.05) / 1.74
> . scalar lambda = -1*normalden(alpha) / normal(alpha)
> . scalar condmean = 2.05 + 1.74*lambda
> . scalar numltone = 10000000*normal(alpha)
> . scalar condtotal = condmean*numltone
> . di condtotal
> -187432.23
>
> Similarly,
>
>   E[X | X > 1] = mu + sigma*lambda'(alpha)
>
> where
>
>   lambda' = phi(alpha) / (1 - Phi(alpha))
>
> In Stata,
>
> . scalar lambda2 = normalden(alpha) / (1 - normal(alpha))
> . scalar condmean2 = 2.05 + 1.74*lambda2
> . scalar condtotal2 = condmean2*(10000000 - numltone)
> . di condtotal2
> 20687432
>
> . sum x if x > 1, mean
> . di r(sum)
> 20685854
>
>
>  -- Brian Poi
>  -- bpoi@stata.com
> *
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> *   http://www.stata.com/help.cgi?search
> *   http://www.stata.com/support/statalist/faq
> *   http://www.ats.ucla.edu/stat/stata/
>



-- 
Steven Samuels
sjsamuels@gmail.com
18 Cantine's Island
Saugerties NY 12477
USA
Voice: 845-246-0774
Fax: 206-202-4783

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